Question

6. A drainage tile is a cylindrical shell 21 cm long. The inside and outside diameters are 8 cm and 10
respectively. What is the volume of clay required for the tile? ​

Answer 1

AnswEr :

• Height of Cylinder ( h ) = 21 cm
• Inner Diameter = 8 cm
• Inner Radius ( r ) = 4 cm
• Outer Diameter = 10 cm
• Outer Radius ( R ) = 5 cm

• Let's Head to the Question Now :

$\implies \tt Volume\: Required = Outer \:Volume - Inner \:Volume \\ \\\implies \tt Volume\: Required = \pi {R}^{2}h - \pi {r}^{2} h \\ \\\implies \tt Volume\: Required = \pi h({R}^{2} - {r}^{2}) \\ \\\implies \tt Volume\: Required = \dfrac{22}{ \cancel{7}} \times \cancel{21} \times ( {5}^{2} - {4}^{2}) \\ \\\implies \tt Volume\: Required =22 \times 3 \times (25 - 16) \\ \\\implies \tt Volume\: Required =66cm\times 9 {cm}^{2} \\ \\\implies \boxed{ \pink{ \tt Volume\: Required =594 \:{cm}^{3} }}$

⠀

∴ Volume of Clay required is 594cm³

$\rule{300}{2}$

$\star \: \underline \text{Some Information about Cylinder :}$

⋆ A cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.

⋆ The area of the curved surface of the cylinder which is contained between the two parallel circular bases. CSA = 2πrh

⋆ The total surface area of a cylinder is the sum of curved surafce area and the area of two circular bases. TSA = 2πr(r + h)

⋆ Volume = πr²h

#answerwithquality #BAL

Answer 2

$\bf{\Huge{\underline{\boxed{\bf{\blue{ANSWER\::}}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}}$

A drainage tile is a cylindrical shell 21cm long. The Inside & outside diameters are 8cm & 10cm respectively.

$\bf{\Large{\underline{\bf{To\:find\::}}}}}$

The volume of clay required for the tile.

$\bf{\Large{\underline{\bf{\orange{Explanation\::}}}}}$

We know that formula of the volume of cylinder:

$\longmapsto{\sf{\pi  r^{2} h}}$   [cubic units]

$\bf{\huge{\underline{\boxed{\sf{\green{First\:case\::}}}}}}$

$\bf{We\:have\begin{cases}\sf{The\:inside\:diameter\:of\:cylinder=8cm}\\ \sf{The\:height\:of\:cylinder=21cm}\end{cases}}$

→ The Inside radius of cylinder = $\sf{\frac{Diameter}{2} }$

→ The inside radius of cylinder = $\sf{\cancel{\frac{8cm}{2}} }$

→ The inside radius of cylinder = 4cm

Therefore,

$\hookrightarrow\sf{Volume\:=\:\pi r^{2} h}$

$\hookrightarrow\sf{Volume\:=\:(\frac{22}{7} *4*4*21)cm^{3}}$

$\hookrightarrow\sf{Volume\:=\:(\frac{22}{\cancel{7}} *4*4*\cancel{21})cm^{3}}$

$\hookrightarrow\sf{volume\:=\:(22*4*4*3)cm^{3}}$

$\hookrightarrow\sf{volume\:=\:1056cm^{3}}$

$\bf{\huge{\underline{\boxed{\sf{\green{Second\:case\::}}}}}}$

$\bf{We\:have\begin{cases}\sf{The\:outside\:diameter\:of\:cylinder=10cm}\\ \sf{The\:height\:of\:cylinder=21cm}\end{cases}}$

→ The outside radius of cylinder = $\sf{\frac{Diameter}{2} }$

→ The outside radius of cylinder = $\sf{\cancel{\frac{10cm}{2}} }$

→ The outside radius of cylinder = 5cm

Therefore,

$\hookrightarrow\sf{Volume\:=\:\pi R^{2} h}$

$\hookrightarrow\sf{Volume\:=\:(\frac{22}{7} *5*5*21)cm^{3}}$

$\hookrightarrow\sf{Volume\:=\:(\frac{22}{\cancel{7}} *5*5*\cancel{21})cm^{3}}$

$\hookrightarrow\sf{volume\:=\:(22*5*5*3)cm^{3}}$

$\hookrightarrow\sf{volume\:=\:1650cm^{3}}$

__________________________________________________

$\bf{\Large\underline{\bf{\pink{The\:volume\:of\:clay\:required\:for\:the \:tile\::}}}}$

→ Outside volume - Inside Volume

→ 1650cm³ - 1056cm³

→ 594cm³

Thus,

The volume of clay is required for the tile is 594cm³.